Question

The figure shows three distance observations $D_1$, $D_2$ and $D_3$. The table lists values of these observations and the corresponding weights. Assuming uncorrelated observations, the most probable values by the least squares approach for these measurements are ............................. (Rounded off to 3 decimal places). \[ \begin{array}{|c|c|c|} \hline \text{Distance} & \text{Measurement (m)} & \text{Weight} \\ \hline D_1 & 40.150 & 1 \\ \hline D_2 & 40.180 & 2 \\ \hline D_3 & 80.390 & 1 \\ \hline \end{array} \]

• A. $\hat D_1=40.170\ \mathrm{m},\ \hat D_2=40.190\ \mathrm{m},\ \hat D_3=80.360\ \mathrm{m}$
• B. $\hat D_1=40.174\ \mathrm{m},\ \hat D_2=40.192\ \mathrm{m},\ \hat D_3=80.366\ \mathrm{m}$
• C. $\hat D_1=40.175\ \mathrm{m},\ \hat D_2=40.185\ \mathrm{m},\ \hat D_3=80.360\ \mathrm{m}$
• D. $\hat D_1=40.172\ \mathrm{m},\ \hat D_2=40.195\ \mathrm{m},\ \hat D_3=80.367\ \mathrm{m}$

Hint:

For a single linear condition $A v=b$, the residuals are $v=W^{-1}A^\top( A W^{-1}A^\top)^{-1} b$. Here $b$ is just the closure error; distribute it according to the inverse weights.

Answer 1

The Correct Option is B

Constraint (from geometry): \( D_1 + D_2 - D_3 = 0 \). Let residuals be \( v_i \) with adjusted \( \hat D_i = D_i + v_i \) and weights \( w = \mathrm{diag}(1,2,1) \). The constraint in residuals: \[ v_1 + v_2 - v_3 = -(D_1 + D_2 - D_3) = -\left( 40.150 + 40.180 - 80.390 \right) = +0.060. \] Minimize \( \Phi = v^\top W v \) subject to \( A v = b \), with \( A = [1\ 1\ -1] \) and \( b = 0.060 \). Constrained least squares gives \[ v = W^{-1} A^\top \left( A W^{-1} A^\top \right)^{-1} b. \] Since \( W^{-1} = \mathrm{diag}(1, \tfrac{1}{2}, 1) \), \[ A W^{-1} A^\top = [1\ 1\ -1] \begin{bmatrix} 1 & 0 & 0 \\ 0 & \tfrac{1}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} = 2.5, \quad (\,.\,)^{-1} = 0.4. \] Thus, \[ v = \begin{bmatrix} 1 \\ \tfrac{1}{2} \\ -1 \end{bmatrix} \left( 0.4 \times 0.060 \right) = \begin{bmatrix} 0.024 \\ 0.012 \\ -0.024 \end{bmatrix} \ \text{m}. \] Therefore (rounded to 3 decimals): \[ \hat D_1 = 40.150 + 0.024 = \mathbf{40.174\ m}, \quad \hat D_2 = 40.180 + 0.012 = \mathbf{40.192\ m}, \quad \hat D_3 = 80.390 - 0.024 = \mathbf{80.366\ m}. \]